Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{3x - 18}{x - 3} \times \dfrac{5x^2 - 35x + 60}{2x^2 - 6x - 36} $
Answer: First factor out any common factors. $n = \dfrac{3(x - 6)}{x - 3} \times \dfrac{5(x^2 - 7x + 12)}{2(x^2 - 3x - 18)} $ Then factor the quadratic expressions. $n = \dfrac {3(x - 6)} {x - 3} \times \dfrac {5(x - 3)(x - 4)} {2(x - 6)(x + 3)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {3(x - 6) \times 5(x - 3)(x - 4) } {(x - 3) \times 2(x - 6)(x + 3) } $ $n = \dfrac {15(x - 3)(x - 4)(x - 6)} {2(x - 6)(x + 3)(x - 3)} $ Notice that $(x - 6)$ and $(x - 3)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {15(x - 3)(x - 4)\cancel{(x - 6)}} {2\cancel{(x - 6)}(x + 3)(x - 3)} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $n = \dfrac {15\cancel{(x - 3)}(x - 4)\cancel{(x - 6)}} {2\cancel{(x - 6)}(x + 3)\cancel{(x - 3)}} $ We are dividing by $x - 3$ , so $x - 3 \neq 0$ Therefore, $x \neq 3$ $n = \dfrac {15(x - 4)} {2(x + 3)} $ $ n = \dfrac{15(x - 4)}{2(x + 3)}; x \neq 6; x \neq 3 $